Size of data after AES/CBC and AES/ECB encryption

Solution 1:

AES has a fixed block size of 16-bytes regardless key size. Assuming you use PKCS 5/7 padding, use this formula,

 cipherLen = (clearLen/16 + 1) * 16;

Please note that if the clear-text is multiple of block size, a whole new block is needed for padding. Say you clear-text is 16 bytes. The cipher-text will take 32 bytes.

You might want to store IV (Initial Vector) with cipher-text. In that case, you need to add 16 more bytes for IV.

Solution 2:

AES, as a block cipher, does not change the size. The input size is always the output size.

But AES, being a block cipher, requires the input to be multiple of block size (16 bytes). For this, padding schemes are used like the popular PKCS5. So the answer is that the size of your encrypted data depends on the padding scheme used. But at the same time all known padding schemes will round up to the next module 16 size (size AES has a 16 bytes block size).

Solution 3:

It depends on the mode in which you use AES. What you have is accurate for most of the block oriented modes, such as ECB and CBC. OTOH, in CFB mode (for one example) you're basically just using AES to produce a stream of bytes, which you XOR with bytes of the input. In this case, the size of the output can remain the size of the input rather than being rounded up to the next block size as you've given above.